# How many ring-LWE samples are required for the (Search) Ring Learning With Errors problem to have a unique solution?

Consider the LWE distribution $$\{(\pmb{a}_{i},\left<\pmb{a}_{i} , \pmb{s}\right> + e_{i})\}$$ where secret $$\pmb{s} \in \mathbb{Z}_{q}^{n}$$, randomness is $$\pmb{a}_{i} \xleftarrow{\} \mathbb{Z}_{q}^{n}, e_{i} \xleftarrow{\} \chi$$, $$q$$ is a prime number and $$\chi$$ is a discrete Gaussian distribution. Grouping $$m$$ samples we obtain $$(\pmb{A},\pmb{A}\pmb{s}+\pmb{e}) \in \mathbb{Z}_{q}^{m \times n}\times \mathbb{Z}_{q}^{m}$$.

We can wonder if the (search) Learning With Errors problem has more than one solution (with bounded error terms). We can compute the probability over the choices of $$\pmb{a}_{i}$$ of the existance of another $$\pmb{s}' \neq \pmb{s}$$ such that $$\pmb{A}(\pmb{s}-\pmb{s}') = \pmb{e}'-\pmb{e}$$ has small norm. If we just want to bound the infinity norm it is easy compute this probability for a particular $$\pmb{s}'$$ and then do a union bound over all possible $$\pmb{s}'$$. As a result we get that $$m\in\mathcal{O}(n)$$ is enough to ensure that the problem has a unique solution except with negligible probability in $$n$$.

I have also read that Chernoff's bounds could be used to analize these probabilities.

Consider now the ring-LWE distribution, $$\{(a_{i},a_{i} \cdot s+e_{i})\}$$, where $$s \in R_{q} = \mathbb{Z}_{q}[x]/\left$$, $$a_{i} \xleftarrow{\} R_{q}$$ and $$e_{i}\in R_{q}$$ is obtained sampling its coefficients from $$\chi$$.

Since we can think of each ring-LWE sample as $$n$$ LWE samples I would expect only a constant number of samples to ensure that the problem of recovering $$s$$ has a unique solution with overwhelming probability. However I do not know how to prove it, since now I cannot use that each $$\pmb{a}_i$$ was independent, and $$R_{q}$$ can have divisors of zero.

Is there a general way of obtaining the number of samples required for this condition? Or general applications only require that the problem is hard but never require uniqueness?